问题
my code is:
public class Box
{
public static void main(String[] args)
{
Integer z = new Integer(43);
z++;
Integer h = new Integer(44);
System.out.println("z == h -> " + (h == z ));
}
}
Output:-
z == h -> false
why the output is false when the values of both the objects is equal?
Is there any other way in which we can make the objects equal?
回答1:
You are trying to compare two different object and not their values. z and h points to two different Integer object which hold same value.
z == h
Will check if two objects are equal. So it will return false.
If you want to compare values stored by Integer object use equals method.
Integer z = new Integer(43); // Object1 is created with value as 43.
z++; // Now object1 holds 44.
Integer h = new Integer(44); // Object2 is created with value as 44.
So at the end we have two different Integer object ie object1 and object2 with value as 44. Now
z = h
This will check if objects pointed by z and h is same. ie object1 == object2
which is false.
If you do
Integer z = new Integer(43); // Object1 is created with value as 43.
z++; // Now object1 holds 44. Z pointing to Object1
Integer h = z; // Now h is pointing to same object as z.
Now
z == h
will return true.
This might help http://www.programmerinterview.com/index.php/java-questions/java-whats-the-difference-between-equals-and/
回答2:
No. Use h.equals(z) instead of h == z to get the equality behavior you expect.
回答3:
h == z would work only if you assign the value via auto-boxing (i.e Integer a = 43) and the value is in between -128 and 127 (cached values), i.e:
Integer a = 44;
Integer b = 44;
System.out.println("a == b -> " + (a == b));
OUTPUT:
a == b -> true
If the value is out of the range [-128, 127], then it returns false
Integer a = 1000;
Integer b = 1000;
System.out.println("a == b -> " + (a == b));
OUTPUT:
a == b -> false
However, the right way to compare two objects is to use Integer.equals() method.
回答4:
Integer is Object not primitive (int) And Object equality compare with equals method.
When you do z == h it will not unbox into int value unless it checks both Integer reference(z & h) are referring same reference or not.
As it is derived in documentation -
The result is true if and only if the argument is not null and is an Integer object that contains the same int value as this object.
System.out.println("z == h -> " + h.equals( z));
It will print true.
回答5:
Integer is an object, not a primitive. If z & h were primitives, == would work just fine. When dealing with objects, the == operator doesn't check for equality; it checks if the two references point to the same object.
As such, use z.equals(h); or h.equals(z); These should return true.
回答6:
Read this: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html
Integer is Object you compare address/references/pointers of objects not values.
Integer a = Integer(1);
Integer b = Integer(1);
a == b; // false
a.compareTo(b); // true
回答7:
check to make sure you can use z++ on the Integer object.
来源：https://stackoverflow.com/questions/14786014/why-2-objects-of-integer-class-in-java-cannot-be-equal